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ScienceCh 11 Electricity

Electricity: Class 10 Science · Chapter 11

Science · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 100 min
  • 3 simulations
  • 46 practice questions
  • 27 MCQs

What you will learn in Electricity

  • Define electric current and potential difference, use I = Q/t and V = W/Q, and say how ammeters and voltmeters are connected
  • Draw and read circuit diagrams with the standard symbols
  • State Ohm's law, check it from V–I readings and use V = IR to solve problems
  • Explain how resistance depends on length, area of cross-section and material, and use R = ρl/A
  • Find the equivalent resistance of resistors in series, in parallel and in simple combinations, and the current and voltage in each
  • Explain the heating effect of current with H = I²Rt and its use in heaters, filament bulbs and fuses
  • Calculate electric power and energy, convert between kilowatt hours and joules, and work out the cost of electricity

Original companion text aligned to NCERT Science Class 10, Chapter 11, pp. 171–194. Not affiliated with or endorsed by NCERT.

Think of the moment the power comes back after a cut on a summer evening. Fans begin to turn, tube lights flicker on, a phone left on charge lights up, and someone checks that the fridge is humming again. Every one of these runs on the same thing: electric charge pushed round a closed path by a source of energy.

In this chapter you will find out what an electric current is, what makes it flow and what controls its size. You will meet Ohm's law and see how resistors behave when they are joined in series and in parallel. Finally you will see why a current heats a wire, how much power appliances use, and how an electricity bill is worked out.

11.1 Electric Current and Circuit

Press a doorbell switch and the bell at the far end of the corridor rings at once. Something has been set moving through the wires: electric charge. A steady flow of charge is an electric current (विद्युत धारा). In metal wires the charges that move are electrons.

Notice that a doorbell needs two wires, not one. Charge leaves the supply, passes through the bell and returns to the supply: it needs a complete route. An unbroken conducting loop from one terminal of a source round to the other is an electric circuit (विद्युत परिपथ). A switch is simply a gap in the loop that you can open or close. Open it, and the flow stops in every part of the loop together; close it, and it starts everywhere together (Fig. 11.1).

3/3By agreement, current is drawn from + to − outside the cell: opposite to the electrons.

Fig. 11.1A switch completes the loop; electrons drift one way, conventional current is drawn the other way.NCERT Fig. 11.1 · p. 172

A single loop of wire joins a cell on the left, a switch at the top, a bulb on the right and an ammeter at the bottom. In the first step the switch is open, the bulb is dark and nothing moves. In the second step the switch is closed, the bulb glows, and electrons, drawn as small blue dots with minus signs, drift from the negative plate of the cell round the loop to the positive plate. In the third step red arrows show the conventional current going the other way, from the positive terminal through the bulb and ammeter back to the negative terminal.

Which way does a current go? The rule we still use was fixed in the early days of electrical science, more than a century before the electron was discovered in 1897: current is taken to leave the positive terminal of the source, go round the circuit and return to the negative terminal. This is called conventional current. Electrons are negative, so in a wire they actually drift the other way, from − to +.

How big is a current? We count how much charge crosses any cross-section of the wire every second. If a charge Q passes in a time t, the current is

I=Qt
(11.1)
where
I
current, in amperes (A)
Q
charge that passes, in coulombs (C)
t
time taken, in seconds (s)

Charge is measured in coulombs (C). One electron carries only 1.6×10−19 C, so a single coulomb is the charge of about 6.25×1018 electrons. The unit of current, the ampere (A), honours André-Marie Ampère (1775–1836) of France. A current of 1 A means one coulomb of charge passing each second, so 1 A=1 C/s. Small currents are given in milliamperes, 1 mA=10−3 A, and microamperes, 1 μA=10−6 A.

Definition
Electric current

The amount of charge that passes a given cross-section of a conductor each second, I=Q/t. Its SI unit is the ampere (A).

ExampleA current of 2 A carries 2 C of charge past a point every second.

Current is measured with an ammeter (ऐमीटर). The current has to pass through the meter to be counted, so an ammeter is always joined in series: you open the circuit and close it again through the meter. In Fig. 11.1 the ammeter sits in the bottom wire. In a single loop its reading would be the same wherever you put it.

Example 11.1NCERT Example 11.1 · p. 172

Charge through a charging cable

A phone charger cable carries a steady current of 1.5 A for 40 minutes. (a) How much charge flows through it? (b) How many electrons carry this charge? (Charge on an electron =1.6×10−19 C.)

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Practice 11.1 (NCERT in-text questions, p. 172): questions with answers

  1. Short answerEasyNCERT in-text Questions p. 172, Q1

    Riya joins a wire from one terminal of a cell to a torch bulb but leaves the bulb's other contact unconnected. The bulb stays dark. Use the idea of an electric circuit to explain why, and say what she must add.

    Show solution

    AnswerThere is no closed path; she must join the bulb's other contact to the other terminal of the cell.

    Charge flows only round a complete, closed conducting path: an electric circuit. Here the path ends at the bulb, so no charge can return to the cell and no current flows. Riya must add a second wire from the bulb's free contact to the other terminal of the cell (a switch in this wire would let her turn the bulb on and off).

  2. Short answerEasyNCERT in-text Questions p. 172, Q2

    Explain exactly what is meant by a current of one ampere, and write the ampere in terms of other units.

    Show solution

    Answer1 A is a flow of 1 C of charge through a cross-section each second: 1 A = 1 C/s.

    Current is the rate of flow of charge, I=Q/t. One ampere is the current when one coulomb of charge crosses any cross-section of the conductor every second, so 1 A=1 C/s.

  3. Quick answerMediumNCERT in-text Questions p. 172, Q3

    A wire carries a current of 1 mA. How many electrons pass a point in the wire each second? (Charge on an electron =1.6×10−19 C.)

    electrons
    Show answer

    Answer 6250000000000000 electrons

    In 1 s the charge is Q=It=10−3 C. The number of electrons is 10−3/(1.6×10−19)=6.25×1015.
Quick check
1 of 3
In the wires outside a cell, which way do the electrons drift?
Show answer

Answer (B) From the negative terminal to the positive terminal

Electrons are negative, so they drift from the negative terminal round to the positive one. The direction from + to − is the conventional current, chosen before electrons were known.

11.2 Electric Potential and Potential Difference

A copper wire is packed with free electrons, yet on its own it carries no current. Something has to push the charges along. Picture water in a closed loop of pipe (Fig. 11.2). If nothing drives it, the water just sits there. Switch on a pump and it creates a difference in pressure: water leaves the pump at high pressure, flows round the loop, and comes back at low pressure to be pushed again.

3/3Cell = pump, bulb = narrow pipe, current = flow, potential difference = pressure gap.

Fig. 11.2A pump makes water flow by keeping up a pressure difference; a cell does the same for charge.

A closed loop of water pipe has a pump on its left side and a narrow section on its right side. In the first step the pump is off and the water does not move. In the second step the pump turns, the pipe just after the pump is marked high pressure and the pipe just before it low pressure, and arrows show water flowing clockwise round the loop and squeezing through the narrow pipe. In the third step the parts are renamed for a circuit: the pump is the cell, the narrow pipe is a bulb or resistor, the flow is the current and the pressure gap is the potential difference.

In a circuit the cell plays the part of the pump. Chemical reactions inside it push charge towards one terminal and away from the other, so the two terminals differ in electrical 'pressure'. That difference is the potential difference (विभवांतर), or p.d. A new cell on a shop shelf already has it. Join a circuit, and the p.d. drives charge round: every coulomb collects energy inside the cell and hands it over in the bulb. The energy comes from the cell's chemicals, which is why cells run down. Two or more cells joined together make a battery.

So p.d. is measured as energy per unit charge. If carrying a charge Q between two points takes work W, the p.d. between those points is

V=WQ
(11.2)
where
V
potential difference, in volts (V)
W
work done, in joules (J)
Q
charge moved, in coulombs (C)

The unit of p.d. is the volt (V), honouring the Italian physicist Alessandro Volta (1745–1827). When 1 J of work moves 1 C of charge between two points, the p.d. between them is one volt:

1 V=1 J1 C=1 J C−1
(11.3)

So a 1.5 V cell hands 1.5 J of energy to every coulomb that passes through it. P.d. is measured with a voltmeter (वोल्टमीटर), which is always joined in parallel: its two leads touch the two points, one on each side of the part being tested, without breaking the circuit.

Definition
Potential difference

The energy transferred per coulomb when charge moves between two points of a circuit, V=W/Q. Its SI unit is the volt; 1 V=1 J/C.

ExampleA 9 V battery gives 9 J to each coulomb that passes through it.

Example 11.2NCERT Example 11.2 · p. 173

Work done by an e-rickshaw battery

An e-rickshaw's battery pack has a potential difference of 48 V. How much work does it do in moving 5 C of charge round the motor circuit?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Practice 11.2 (NCERT in-text questions, p. 174): questions with answers

  1. Quick answerEasyNCERT in-text Questions p. 174, Q1

    In a torch, which part supplies the push (potential difference) that drives charge through the bulb's filament?

    Show answer

    Answer (D) the cells

    The cells keep a p.d. between their terminals by chemical action. The switch and case only complete the path.
  2. Short answerEasyNCERT in-text Questions p. 174, Q2

    The potential difference between the two ends of a wire is 1 V. What does this tell you about energy and charge?

    Show solution

    Answer1 J of work is done for every 1 C of charge moved from one end to the other.

    From V=W/Q, a p.d. of 1 V means that 1 J of work is done (energy transferred) for each coulomb of charge that moves between the two ends.

  3. NumericalEasyNCERT in-text Questions p. 174, Q3

    A car battery has a p.d. of 12 V.

    1. How much energy does each coulomb of charge gain as it passes through the battery?

      J
      Show answer

      Answer 12 J

      W=VQ=12×1=12 J.
    2. How much energy do 3 C of charge gain?

      J
      Show answer

      Answer 36 J

      W=VQ=12×3=36 J.
Quick check
1 of 3
In the water-loop picture of a circuit, what plays the part of the potential difference?
Show answer

Answer (D) The pressure difference made by the pump

The pump keeps a pressure difference that drives the flow, just as a cell keeps a p.d. that drives the current. The water is like the charge and the narrow pipe like a resistance.

11.3 Circuit Diagram

Sketching a circuit as it really looks takes time and hides what matters. Instead we draw a circuit diagram: each part becomes a simple standard symbol, and wires become straight lines. Anyone who knows the symbols can then read the circuit, whatever language they speak. Fig. 11.3 shows the symbols used in this chapter.

1/5The symbols you will use in this chapter.

Fig. 11.3Standard symbols let anyone read a circuit at a glance.NCERT Table 11.1 · p. 174

Twelve circuit symbols in two columns, each with its name. The left column shows a cell, a battery of two cells, an open switch, a closed switch, two wires joined with a dot, and two wires crossing without joining, where one wire hops over the other. The right column shows a bulb as a circle with a cross, a resistor as a zigzag, a rheostat as a zigzag with an arrow through it, a fuse as a small rectangle on a wire, an ammeter as a circle with A and a voltmeter as a circle with V. The steps highlight the sources, then the switches and wires, then the components, then the meters.

Table 11.1How to recognise each symbol
ComponentWhat the symbol looks like
Cella long thin line (+) beside a short thick line (−)
Batterytwo or more cell symbols joined in a row
Switch or plug keya gap with a lever: open (raised) or closed (flat)
Joined wireslines that meet with a solid dot
Unjoined wireslines that cross with no dot (or with a small hop)
Bulba circle with a cross inside
Resistora zigzag line
Rheostata zigzag with an arrow across it
Fusea small rectangle with the wire running through it
Ammetera circle with the letter A
Voltmetera circle with the letter V

A few habits make diagrams easy to read. Draw wires as straight lines with square corners. Make the longer plate of a cell the positive terminal. Put a dot wherever wires are joined, and leave it out where they only cross. In school labs a plug key often does the switch's job: pushing its brass plug in closes the circuit, and taking it out opens it.

Quick check
1 of 2
Match each symbol with the part it stands for.

Tap an item on the left, then its match on the right.

Show answer
Answer
  • a long thin line beside a short thick line : cell
  • a zigzag line : resistor
  • a zigzag with an arrow across it : rheostat
  • a circle with the letter V : voltmeter
  • two lines meeting at a dot : joined wires
These are the standard symbols in Fig. 11.3. The arrow across a rheostat shows that its resistance can be changed.

11.4 Ohm's Law

If you double the push, do you double the current? To find out, we need a circuit in which the p.d. can be raised step by step while the current is read each time.

Plotting the readings gives a straight line through the origin (Fig. 11.4). Twice the p.d. gives twice the current, three times the p.d. three times the current, and the ratio V/I never changes.

3/4Wire B gives a steeper line: twice the slope, so twice the resistance.

Fig. 11.4For a resistor at fixed temperature, V–I is a straight line through the origin; its slope is R.NCERT Fig. 11.3 · p. 176

A graph with current I in amperes across and potential difference V in volts upwards. In the first step four blue points for wire A lie at 0.3, 0.6, 0.9 and 1.2 amperes with 1.5, 3, 4.5 and 6 volts. In the second step a straight line through the origin joins them, and a slope triangle shows 3 volts divided by 0.6 amperes, so R is 5 ohms. In the third step an orange dashed line for wire B, with points at half the current, is steeper and is labelled 10 ohms. In the fourth step a dotted purple curve for a filament lamp bends upwards, because the lamp's resistance grows as it heats.

In 1827 Georg Simon Ohm (1787–1854), a German physicist, reported this pattern for metal wires. It is now called Ohm's law (ओम का नियम).

LawOhm's law

For a metal conductor kept at a fixed temperature, the current through it is proportional to the potential difference between its ends.

Ohm's law
V=IR
(11.5)
where
R
resistance of the conductor, in ohms (Ω)

The constant R tells you how many volts it takes to push each ampere through the conductor. It is called the resistance (प्रतिरोध): the bigger it is, the more strongly the conductor opposes the flow of charge. It is measured in ohms, written Ω (the Greek capital omega). A wire in which every volt drives exactly one ampere has R=1 Ω, so 1 Ω=1 V/1 A. Ohm's law can be turned round in two useful ways:

R=VI
(11.6)
I=VR
(11.7)
Definition
Resistance

The opposition a conductor offers to the flow of charge, R=V/I. Its SI unit is the ohm (Ω).

ExampleA bulb that draws 0.5 A at 6 V has R=12 Ω.

A dimmer for a stage light, the speed control of a toy car and an old fan regulator all do the same job: they change the current while the supply stays the same. Form (11.7) shows how. With V fixed, current and resistance are inversely proportional, so three times the resistance lets through a third of the current. A part whose resistance can be adjusted is a variable resistor. The lab version, a rheostat (धारा नियंत्रक), has a sliding contact that sets how much resistance wire is in the circuit.

The simulation below repeats the activity. Note one difference: it plots current upwards against p.d., so there the steeper line belongs to the smaller resistance (its slope is 1/R).

SimulationOhm's law bench

A cell with a voltage slider drives current through one of two resistance coils, coil A of 10 ohms or coil B of 20 ohms, with an ammeter in series and a voltmeter across the coil. Each slider setting adds a point to a graph of current against voltage, and a readout gives V divided by I. A switch opens and closes the circuit.

Fig. 11.5Move the Cell slider in steps of 1.5 V to copy Activity 11.1, then switch to coil B.NCERT Fig. 11.2 · p. 175
Try this
  • Set the cell to 1.5, 3, 4.5 and 6 V for coil A. Does the V ÷ I readout change?
  • Switch to coil B and repeat. Which line is steeper on this graph, and why does that mean a smaller resistance?
  • Open the switch. What do the meters read?

Does every part of a circuit pass the same current from the same source? The next activity compares several.

Why do parts differ? Inside a metal the free electrons cannot glide through: they keep colliding with the vibrating atoms, which slows their drift. How much a part slows the drift is its resistance. The rubber band sits at one extreme: an insulator, with so much resistance that no current can be measured. At the other extreme is a good conductor such as copper, whose resistance is tiny for its size. In between are parts made on purpose with a known, useful resistance, like the nichrome wire and the coil you used: these are resistors.

Quick check
1 of 3
A geyser element draws 10 A from a 220 V supply. What is its resistance?
Ω
Show answer

Answer 22 Ω

R=V/I=220/10=22 Ω.

11.5 Factors on which the Resistance of a Conductor Depends

An electrician uses thick wire for an air conditioner but thin wire for a bedside lamp, and heater coils are never made of copper. These choices depend on how the resistance of a wire changes with its size and with the material it is made of.

3/4Twice the area: half the resistance, twice the current.

Fig. 11.6Same cell each time: length, area and material each change the current.NCERT Fig. 11.5 · p. 178

A wire held between two clips is drawn above a horizontal bar whose length shows the current from the same cell. In the first step a constantan wire of length l gives 0.40 amperes. In the second step the wire is twice as long and the bar shrinks to half, 0.20 amperes. In the third step the wire is thicker, with twice the area, and the bar doubles to 0.80 amperes. In the fourth step the wire is copper of the original size, and the bar runs off the scale, labelled far bigger.

Careful measurements on uniform metal wires give two rules. Resistance rises in step with length, R∝l, and falls as the cross-section grows, R∝1/A. A longer wire is a longer obstacle course for the electrons; a thicker wire offers more lanes side by side. Putting the two together gives

R=ρ lA
(11.10)
where
ρ
resistivity of the material, in ohm metres (Ω m)
l
length of the conductor, in metres
A
area of cross-section, in square metres

The constant ρ (Greek rho) is the resistivity (प्रतिरोधकता) of the material. Resistance belongs to a particular wire, but resistivity belongs to the substance. Cut a copper wire in half and its resistance halves, yet its resistivity is unchanged. Rearranging gives ρ=RA/l, so the unit of resistivity is Ω×m2/m, the ohm metre (Ω m).

Definition
Resistivity

The resistance of a conductor of the material that has unit length and unit area of cross-section, ρ=RA/l. It depends on the material and its temperature, not on the wire's shape. SI unit: ohm metre (Ω m).

Table 11.5Resistivity of some metals, alloys and insulators at 20 °C (handbook values for problems; no need to memorise them)NCERT Table 11.2 · p. 179
MaterialResistivity (Ω m)
Silver1.59×10−8
Copper1.68×10−8
Gold2.44×10−8
Aluminium2.65×10−8
Tungsten5.60×10−8
Nickel6.99×10−8
Iron9.7×10−8
Manganin (alloy: Cu, Mn, Ni)4.8×10−7
Constantan (alloy: Cu, Ni)4.9×10−7
Nichrome (alloy: Ni, Cr, Fe)1.10×10−6
Glass (insulator)1011 to 1015
Hard rubber (insulator)about 1013
Fused quartz (insulator)about 7.5×1017

Look at the spread in Table 11.5. Metals and alloys lie between about 10−8 and 10−6 Ω m; insulators such as glass, rubber and quartz lie far above, from about 1011 to beyond 1017 Ω m. Resistivity also changes with temperature: for metals it rises as they get hotter.

These numbers explain why each job gets its own material:

  • Wires and power lines should waste as little energy as possible, so they use the metals of lowest resistivity that are affordable: copper, and aluminium, which is also light.
  • Heating elements in irons, toasters and room heaters need a large resistance in a small coil and must stay red-hot for hours. Alloys such as nichrome suit them: an alloy usually has a far higher resistivity than the metals it is made from, and nichrome does not oxidise (burn) easily.
  • Bulb filaments must glow white-hot without melting, so they are made of tungsten, which survives temperatures above 3000 ∘C.
Example 11.3NCERT Example 11.3 · p. 179

Soldering iron or immersion rod?

On a 220 V supply, how much current is drawn by (a) a soldering iron whose element has a resistance of 880 Ω, and (b) an immersion water-heater rod of resistance 44 Ω?

Try it yourself first. Then reveal the solution one step at a time.

Example 11.4NCERT Example 11.4 · p. 180

Turning up the voltage

A resistor in a toy car draws 0.30 A when the p.d. across it is 4.5 V. What current will it draw if the p.d. is raised to 7.5 V? Assume its temperature does not change.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.5NCERT Example 11.5 · p. 180

Which alloy is this?

Rohan unwinds the heating wire from an old hair dryer. A 2.0 m length has a diameter of 0.40 mm and a resistance of 17.5 Ω at 20 °C. Find the resistivity of the wire and use Table 11.5 to name the material.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.6NCERT Example 11.6 · p. 180

Longer and thinner

A wire of a certain alloy has a resistance of 2 Ω. A second wire of the same alloy is three times as long and has half the area of cross-section. What is its resistance?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 11.3 (NCERT in-text questions, p. 181): questions with answers

  1. Short answerEasyNCERT in-text Questions p. 181, Q1

    Name the three things that decide the resistance of a metal wire at a fixed temperature, and say how each affects it.

    Show solution

    AnswerLength (R ∝ l), area of cross-section (R ∝ 1/A) and the material (through its resistivity ρ).

    From R=ρl/A: a longer wire has more resistance (R∝l); a thicker wire, with a larger area of cross-section, has less (R∝1/A); and the material matters through its resistivity ρ (nichrome far more than copper). Temperature also changes ρ.

  2. Quick answerEasyNCERT in-text Questions p. 181, Q2

    Two copper wires of equal length, one thick and one thin, are joined in turn to the same cell. Through which does the current flow more easily?

    Show answer

    Answer (B) The thick wire, because its area of cross-section is larger

    R∝1/A: the thick wire has the smaller resistance, so a larger current flows through it from the same cell.
  3. Quick answerEasyNCERT in-text Questions p. 181, Q3

    A resistor carries 0.6 A. The p.d. across it is cut to one third of its earlier value; its resistance does not change. What is the new current?

    A
    Show answer

    Answer 0.2 A

    I=V/R, so one third of the p.d. gives one third of the current: 0.6/3=0.2 A.
  4. Short answerEasyNCERT in-text Questions p. 181, Q4

    Why is the coil of an electric room heater made of nichrome and not of copper?

    Show solution

    AnswerNichrome has a high resistivity, so a short coil heats strongly, and it does not oxidise or melt at red heat; copper has a low resistivity and oxidises when hot.

    A heating coil needs a large resistance in a small space, so it needs a material of high resistivity: nichrome's is about 65 times copper's. It must also stay red-hot for hours: nichrome does not oxidise (burn) readily and has a high melting point, while copper would oxidise and pass a dangerously large current.

  5. Quick answerEasyNCERT in-text Questions p. 181, Q5

    Use Table 11.5 to answer these.

    1. Which is the better conductor, nickel or iron?

      Show answer

      Answer (A) nickel

      Nickel's resistivity (6.99×10−8 Ω m) is lower than iron's (9.7×10−8 Ω m).
    2. Which alloy in the table is the poorest conductor?

      Show answer

      Answer (C) nichrome

      Nichrome has the highest resistivity of the alloys listed, 1.10×10−6 Ω m.
Quick check
1 of 3
A wire is replaced by one of the same metal that is twice as long and has twice the area of cross-section. Its resistance:
Show answer

Answer (C) stays the same

R=ρl/A: doubling l doubles R, doubling A halves it, so the two cancel. Choosing four times means both changes were wrongly taken to increase R.

11.6 Resistance of a System of Resistors

Real gadgets rarely hold just one resistor. A radio, a phone charger or the wiring of a house joins many parts together, and there are two basic ways of doing it. In series (श्रेणीक्रम) the resistors are joined end to end in one chain, so charge must pass through each in turn. In parallel (पार्श्वक्रम) every resistor is joined across the same two points, so the charge has a choice of routes. In each case Ohm's law lets us replace the whole group by a single equivalent resistance.

11.6.1 Resistors in Series

Does a resistor 'use up' some of the current, leaving less for the next one? An ammeter moved along a series chain answers this.

2/30.75+1.5+2.25=4.5 V: the p.d.s across the parts add up to the total.

Fig. 11.7In series: one current, shared voltage, and the resistances add.NCERT Fig. 11.8 · p. 183

A rectangular circuit has a 4.5 volt cell in the bottom wire and three zigzag resistors of 5, 10 and 15 ohms along the top wire. In the first step red labels at several places round the loop all read 0.15 amperes, showing the same current everywhere. In the second step brackets over the resistors show 0.75, 1.5 and 2.25 volts, and a larger bracket over all three shows 4.5 volts in total. In the third step the three resistors are replaced by one long resistor labelled 30 ohms, and the current is still 0.15 amperes.

DerivationEquivalent resistance in series

Let the same current I flow through R1, R2 and R3, and let the p.d. across the whole chain be V. The activity showed that the separate p.d.s add up:

V=V1+V2+V3
(11.11)

Ohm's law applies to the single resistor Rs that could replace the chain, keeping the same V and I:

V=IRs
(11.12)

It also applies to each resistor on its own:

V1=IR1V2=IR2V3=IR3
(11.13)

Put all of these into (11.11) and divide every term by I:

IRs=IR1+IR2+IR3
Rs=R1+R2+R3
(11.14)

The equivalent resistance of resistors in series is the sum of their resistances, so it is always larger than the largest of them.

Example 11.7NCERT Example 11.7 · p. 184

A lamp with a dimming resistor

A 15 Ω lamp and a 5 Ω resistor are joined in series to a 9 V battery. Find (a) the total resistance, (b) the current, and (c) the p.d. across the lamp and across the resistor.

Try it yourself first. Then reveal the solution one step at a time.

SimulationSeries circuit explorer

A 4.5 volt cell drives three resistors of 5, 10 and 15 ohms joined in series, with an ammeter and a voltmeter. Readouts show the equivalent resistance, the current and the voltmeter reading. The learner chooses where the voltmeter is connected, selects a resistor to change its value with a slider, changes the cell voltage and opens or closes the switch; moving markers show the current.

Fig. 11.8The chain from Activities 11.4 and 11.5. Move the voltmeter from part to part and change the resistances.
Try this
  • Put the voltmeter across each resistor in turn. Do the three readings add up to the cell's p.d.?
  • Make R2 much larger. What happens to the current, and which resistor now takes most of the voltage?
  • Open the switch. What happens everywhere in the loop at once?
Practice · 2 questions

Practice 11.4 (NCERT in-text questions, p. 185): questions with answers

  1. DiagramEasyNCERT in-text Questions p. 185, Q1

    Draw the circuit diagram for a battery of four 1.5 V cells, a switch, and resistors of 2 Ω, 4 Ω and 6 Ω, all joined in series.

    Show solution

    AnswerOne closed loop: four cell symbols joined + to −, then the switch, then three zigzags labelled 2 Ω, 4 Ω and 6 Ω.

    Draw one rectangular loop. On one side put four cell symbols in a row, each long plate (+) joined to the next cell's short plate (−); label the battery 6 V. Along the other sides put a switch and three zigzag resistors labelled 2 Ω, 4 Ω and 6 Ω, one after another. There are no branches, so no junction dots are needed.

  2. NumericalMediumNCERT in-text Questions p. 185, Q2

    Add two meters to the circuit of Q1: one that reads the current in the chain, and one that reads the p.d. between the two ends of the 6 Ω resistor.

    1. Where do the two meters go?

      Show solution

      AnswerAmmeter anywhere in the loop, in series; voltmeter across the two ends of the 6 Ω resistor, in parallel.

      The ammeter is placed in series anywhere in the loop (the current is the same everywhere). The voltmeter is joined in parallel, one lead at each end of the 6 Ω resistor.

    2. What does the ammeter read?

      A
      Show answer

      Answer 0.5 A

      Rs=2+4+6=12 Ω and V=4×1.5=6 V, so I=6/12=0.5 A.
    3. What does the voltmeter read?

      V
      Show answer

      Answer 3 V

      V=IR=0.5×6=3 V.
Quick check
1 of 2
Resistors of 3 Ω, 4 Ω and 5 Ω are joined in series across a 6 V battery. What current flows?
A
Show answer

Answer 0.5 A

Rs=3+4+5=12 Ω, so I=6/12=0.5 A.

11.6.2 Resistors in Parallel

Now join the resistors side by side, each one connected between the same two points X and Y. Every resistor gets its own branch. The charge arriving at X splits up among the branches and joins up again at Y.

2/30.3+0.2+0.1=0.6 A: the smallest resistor carries the largest current.

Fig. 11.9In parallel: one voltage, currents add, and the equivalent resistance is smaller than any branch.NCERT Fig. 11.10 · p. 185

Three zigzag resistors of 10, 15 and 30 ohms sit on three rungs of a ladder between a left and a right wire, with a 3 volt cell in the bottom wire. In the first step each rung is labelled 3 volts, showing the same p.d. across every resistor. In the second step red arrows on the rungs show currents of 0.3, 0.2 and 0.1 amperes, and an arrow on the left wire shows the main current of 0.6 amperes. In the third step the three rungs are replaced by a single 5 ohm resistor that draws the same 0.6 amperes.

DerivationEquivalent resistance in parallel

The activity showed that the main current is the sum of the branch currents:

I=I1+I2+I3
(11.15)

Every branch has the same p.d. V. Apply Ohm's law to the single resistor Rp that would draw the same main current:

I=VRp
(11.16)

and to each branch on its own:

I1=VR1I2=VR2I3=VR3
(11.17)

Put these into (11.15):

VRp=VR1+VR2+VR3

Divide every term by V:

1Rp=1R1+1R2+1R3
(11.18)

In words: add up the reciprocals of the resistances, and the total is the reciprocal of Rp. Each extra branch gives the charge another route, so Rp is always smaller than the smallest resistor in the group. For just two resistors this becomes Rp=R1R2/(R1+R2), 'product over sum'.

Example 11.8NCERT Example 11.8 · p. 186

Three branches on one battery

Resistors of 4 Ω, 6 Ω and 12 Ω are joined in parallel across a 6 V battery. (a) How much current flows in each branch? (b) How much does the battery supply in all? (c) What single resistor could replace the three?

Try it yourself first. Then reveal the solution one step at a time.

Fig. 11.10Group 1 (6 Ω ∥ 3 Ω) in series with group 2 (12 Ω ∥ 4 Ω).NCERT Fig. 11.12 · p. 187

Circuit diagram. On the left, group 1 is a 6 ohm and a 3 ohm resistor side by side between two junctions. On the right, group 2 is a 12 ohm and a 4 ohm resistor side by side. A wire joins the two groups, and a 10 volt battery in the bottom wire completes the loop.

Example 11.9NCERT Example 11.9 · p. 187

Two parallel groups in series

In Fig. 11.10, a 6 Ω and a 3 Ω resistor in parallel (group 1) are joined in series with a 12 Ω and a 4 Ω resistor in parallel (group 2). A 10 V battery drives the circuit. Find (a) the total resistance and (b) the current drawn from the battery.

Try it yourself first. Then reveal the solution one step at a time.

Why is nothing in your home wired in series? Picture the five lamps along a school corridor joined in one series chain. They would have to share the 220 V, so none would get the voltage it was made for. If one lamp burnt out, the only path would break and the whole corridor would go dark (Fig. 11.11). And a lamp could never share a chain with a heater, because the two need very different currents while a series chain forces the same current through both.

2/3The series row goes dark; the parallel row stays lit.

Fig. 11.11One dead lamp darkens a whole series chain but not a parallel row.NCERT p. 187

Two rows of five lamps. The top row is wired in series, all five on one wire in a single loop with the supply. The bottom row is wired in parallel, each lamp on its own rung between two supply wires. In the first step all ten lamps glow. In the second step the middle lamp of each row is dead: every lamp in the series row goes dark, while the other four lamps in the parallel row stay lit. In the third step arrows show current still reaching each working parallel lamp along its own path, while the series loop is broken at one point.

In parallel, each appliance sits directly across the supply, so each gets the full voltage and draws only the current it needs. Each has its own switch, and one failure leaves the rest working. Joining more appliances in parallel lowers the total resistance and raises the total current drawn from the supply. That is why house wiring needs limits on current, and fuses.

SimulationParallel circuit explorer

A 3 volt cell drives three resistors of 10, 15 and 30 ohms joined in parallel, with an ammeter in the main line and a voltmeter that can be placed across the cell or across the group of resistors. Readouts show the equivalent resistance, the main current, the voltmeter reading and the current in the selected resistor; moving markers show the current splitting into the branches. Sliders change each resistance and the cell voltage, and a switch opens the circuit.

Fig. 11.12The circuit from Activity 11.6. Tap a resistor to read its current, change the resistances and watch how the current divides.
Try this
  • Tap R1, R2 and R3 in turn and note the current in each. Do the three add up to the ammeter reading?
  • Make R3 very large, say 1000 Ω. Does the main current change much? Compare the equivalent resistance with R1 and R2 alone.
  • Make R1 smaller and watch the main current. Which branch now carries the most?
Practice · 5 questions

Practice 11.5 (NCERT in-text questions, p. 188): questions with answers

  1. Short answerMediumNCERT in-text Questions p. 188, Q1

    Without exact working, estimate the equivalent resistance when these are joined in parallel.

    1. 5 Ω and 105 Ω

      Show solution

      AnswerJust under 5 Ω (about 4.9998 Ω).

      Rp must be less than 5 Ω, but the huge 105 Ω branch carries almost no current, so Rp is only a hair below 5 Ω (about 4.9998 Ω).

    2. 5 Ω, 500 Ω and 105 Ω

      Show solution

      AnswerA little under 5 Ω (about 4.95 Ω).

      Again below 5 Ω; the 500 Ω branch takes about 1 % as much current as the 5 Ω one, so Rp is about 1 % lower: ≈4.95 Ω.

  2. NumericalMediumNCERT in-text Questions p. 188, Q2

    A tube light of resistance 880 Ω, a fan of 440 Ω and a TV of 220 Ω run in parallel on a 220 V supply (treat each as a simple resistance). On the same supply, a geyser on its own draws a current equal to the total for all three. Find that current and the geyser's resistance.

    Show solution

    Answer1.75 A; about 126 Ω.

    Each has 220 V across it: I=0.25+0.5+1.0=1.75 A.

    The geyser's resistance is R=V/I=220/1.75≈126 Ω (the equivalent resistance of the three).

  3. Short answerEasyNCERT in-text Questions p. 188, Q3

    Give three advantages of joining the appliances in a house in parallel rather than in series.

    Show solution

    AnswerEach gets the full mains voltage; each can be switched on or off separately; one fault does not stop the others (and each draws only the current it needs).

    (i) Every appliance has the full supply voltage across it, so each works as designed. (ii) Each can have its own switch and be used independently. (iii) If one fails, the others keep working. Also, each draws the current it needs; in series they would all be forced to carry the same current.

  4. DiagramMediumNCERT in-text Questions p. 188, Q4

    You have resistors of 3 Ω, 6 Ω and 9 Ω. Show how to join all three to give a total of:

    1. 11 Ω

      Show solution

      Answer9 Ω in series with (3 Ω ∥ 6 Ω).

      3 Ω∥6 Ω=18/9=2 Ω; then 9+2=11 Ω.

    2. 2.5 Ω

      Show solution

      Answer(6 Ω + 9 Ω) in parallel with 3 Ω.

      6+9=15 Ω in series; then 15∥3=45/18=2.5 Ω.

  5. NumericalMediumNCERT in-text Questions p. 188, Q5

    Four coils have resistances of 5 Ω, 10 Ω, 15 Ω and 30 Ω. Using all four:

    1. What is the largest equivalent resistance possible?

      Ω
      Show answer

      Answer 60 Ω

      All in series: 5+10+15+30=60 Ω.
    2. What is the smallest?

      Ω
      Show answer

      Answer 2.5 Ω

      All in parallel: 1/Rp=(6+3+2+1)/30=12/30, so Rp=2.5 Ω.
Quick check
1 of 3
Two 10 Ω resistors are joined in parallel. The equivalent resistance is:
Show answer

Answer (C) 5 Ω

1/Rp=1/10+1/10=2/10, so Rp=5 Ω. 0.2 is the value of 1/Rp: a common slip is to forget to turn the fraction over.

11.7 Heating Effect of Electric Current

Touch a phone charger after an hour of use: it is warm. A mixer-grinder turns most of its energy into the spinning of its blades, yet its body warms up too. A geyser element is nothing but a resistor, so every joule it receives becomes heat. Whenever a current flows through a conductor that has resistance, some or all of the source's energy is turned into heat. This is the heating effect of electric current.

DerivationHow much heat does a current produce?

Let a steady current I flow for a time t through a resistor R that has a p.d. V across it. In that time a charge Q=It passes through. Moving a charge Q through a p.d. V takes work VQ, so the source supplies energy at the rate

P=VQt=VI
(11.19)

In time t the source supplies energy VIt. In a purely resistive circuit all of it appears as heat H in the resistor:

H=VIt
(11.20)

Use Ohm's law, V=IR, to write this in terms of the current:

H=I2Rt
(11.21)
LawJoule's law of heating

The heat produced in a resistor is H=I2Rt: for a given resistor it grows as the square of the current; for a given current it is proportional to the resistance; and it is proportional to the time the current flows.

Notice the square in Joule's law. Doubling the current through the same resistor produces four times the heat, while doubling the resistance or the time only doubles it (Fig. 11.13). For an appliance on a known supply voltage, first find the current from I=V/R, then use H=I2Rt; or use H=V2t/R directly.

4/4Twice the current: (2I)2=4I2, so 4H.

Fig. 11.13H=I2Rt: doubling t or R doubles the heat, doubling I makes it four times.NCERT p. 189

Four columns of heat blocks stand on a baseline. The first column, for current I, resistance R and time t, has one block labelled H. In the second step a column for twice the time appears with two blocks, 2H. In the third step a column for twice the resistance also has two blocks, 2H. In the fourth step a column for twice the current has four blocks, 4H, the tallest of all.

Example 11.10NCERT Example 11.10 · p. 189

Hot and warm settings of a hair dryer

A hair dryer on a 220 V supply uses 1100 W on its hot setting and 550 W on its warm setting. Find the current and the resistance of the heating element in each setting.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.11NCERT Example 11.11 · p. 189

Finding the voltage from the heat

A 10 Ω resistor produces 360 J of heat in 4 s. What is the p.d. across it?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
The current through a heater element is doubled while its resistance stays the same. The heat it produces each second becomes:
Show answer

Answer (B) four times as much

H=I2Rt, so doubling I multiplies the heat by 22=4. 'Twice' forgets the square.

11.7.1 Practical Applications of Heating Effect of Electric Current

Heat from a current is sometimes a nuisance. It wastes energy in power lines and chargers, and the extra warmth can change how components behave or even damage them. But we also use it on purpose every day: in irons, toasters, ovens, kettles, geysers and room heaters, a high-resistance element turns electrical energy into heat.

An old-style filament bulb is really a heater that happens to glow: more than 90 % of the energy it uses comes out as heat, not light. To glow white, its filament must pass 2500 ∘C, so it is made of tungsten, which melts only at about 3400 ∘C. Two design tricks help. Thin supports keep heat from leaking away into the base, and the glass is filled with unreactive nitrogen and argon instead of air, so the white-hot metal cannot burn.

Joule heating also keeps us safe. Suppose too many heaters are switched on together, or the two wires in a damaged cord touch. The current rises far above normal, and the house wiring itself could overheat and start a fire. A fuse (फ़्यूज़) prevents this. It is a deliberately weak link: a short, thin wire with a fairly low melting point, placed in series so that the whole current passes through it. Because heating grows as I2, the fuse wire melts within moments once the current passes its rating, and the circuit is broken before anything else gets dangerously hot (Fig. 11.14). The wire usually sits in a porcelain or glass cartridge with metal end caps.

2/3Iron, heater and kettle together draw about 20 A: the fuse wire heats up fast.

Fig. 11.14A fuse melts and opens the circuit before too large a current can do damage.NCERT p. 190

The live wire runs along the top and the neutral wire along the bottom. A fuse cartridge marked 10 A sits in the live wire, before the three appliances, an iron, a heater and a kettle, which are joined in parallel between the wires. In the first step only the iron is on and the current is 4.5 amperes, which is safe. In the second step the heater and kettle are switched on too, the current reaches 20 amperes and the fuse wire glows hot. In the third step the fuse wire has melted, leaving a gap, so the current is zero and all three appliances are off.

Household fuses are made for set currents, such as 2 A, 5 A, 10 A or 15 A. Choose a rating a little above the current the appliance normally draws. For example, a 2 kW room heater on 220 V draws I=P/V=2000/220≈9.1 A, so it needs a 10 A fuse. A 5 A fuse would melt the moment the heater was switched on, while a 15 A one would give it less protection. Many modern homes use a miniature circuit breaker (MCB) instead: a switch that trips on overload and can simply be reset.

Practice · 3 questions

Practice 11.6 (NCERT in-text questions, p. 190): questions with answers

  1. Short answerMediumNCERT in-text Questions p. 190, Q1

    An electric kettle's flexible cable and its heating element carry the same current. Why does the element get very hot while the cable stays nearly cool?

    Show solution

    AnswerHeat H = I²Rt; with the same current, the high-resistance element produces far more heat than the low-resistance copper cable.

    Both carry the same current for the same time, so by H=I2Rt the heat produced in each is proportional to its resistance. The element is nichrome with a large resistance; the cable is thick copper with a tiny resistance, so almost all the heat appears in the element.

  2. Quick answerMediumNCERT in-text Questions p. 190, Q2

    A charge of 18 000 C flows through a heater in 2 hours. The p.d. across the heater is 24 V. How much heat is produced?

    J
    Show answer

    Answer 432000 J

    H=VQ=24×18 000=4.32×105 J. The time is not needed once you know the charge.
  3. Quick answerEasyNCERT in-text Questions p. 190, Q3

    A kettle element of resistance 40 Ω carries 4 A. How much heat does it produce in 2 minutes?

    J
    Show answer

    Answer 76800 J

    H=I2Rt=42×40×120=76 800 J.
Quick check
1 of 3
How is a fuse connected?
Show answer

Answer (A) In series with the circuit it protects

A fuse must carry the same current as the circuit so that it melts when that current is too large: it goes in series. Joined across the supply, it would itself be a short circuit.

11.8 Electric Power

Why does a geyser push up a bill so much faster than a ceiling fan? Because it turns electrical energy into heat far more quickly. The rate at which an appliance converts electrical energy into heat, light, sound or motion is its electric power (विद्युत शक्ति). You met it in (11.19) as P=VI; Ohm's law gives two more forms:

Electric power
P=VI=I2R=V2R
(11.22)

Use P=I2R when you know the current, as in series circuits, and P=V2/R when you know the voltage, as for appliances on the mains. The SI unit of power is the watt (W): one watt is used by a device that carries 1 A under a p.d. of 1 V.

1 W=1 V×1 A=1 V A
(11.23)
Definition
Electric power

The rate at which electrical energy is used, P=VI=I2R=V2/R. Its SI unit is the watt (W).

ExampleA 220 V fan drawing 0.25 A uses 55 W.

Appliance ratings are often given in kilowatts, since 1 kW=1000 W. Your electricity meter does not count joules: a joule is far too small for a whole month's use. Because energy = power × time, it counts power in kilowatts multiplied by time in hours. This unit, the kilowatt hour (kW h), is what your bill calls a unit. (A smaller relative, the watt hour, W h, is the energy a 1 W device uses in an hour.)

A kilowatt is 1000 J every second and an hour is 3600 s, so one kilowatt hour is 1000×3600=3 600 000 joules:

1 kW h=3.6×106 J
Definition
Kilowatt hour (one unit)

The energy used by a device of power 1 kW in one hour: 1 kW h=3.6×106 J. It is the commercial unit of electrical energy.

ExampleA 100 W bulb left on for 10 hours uses 1 unit.

4/4Each rectangle has the same area: one unit of energy.

Fig. 11.15Power × time = energy, so all three rectangles are the same one unit.NCERT p. 191

A graph with time in hours across and power in kilowatts upwards, so the area of a rectangle is energy. In the first step a square shows a 1 kilowatt heater used for 1 hour. In the second step a long thin strip shows a 100 watt bulb used for 10 hours. In the third step a tall narrow column shows a 2 kilowatt geyser used for 30 minutes. In the fourth step all three are shown together with the note that each area equals 1 kilowatt hour, one unit.

Example 11.12NCERT Example 11.12 · p. 191

Power of a mixer-grinder

A mixer-grinder draws 2.5 A from a 220 V supply. What is its power?

Try it yourself first. Then reveal the solution one step at a time.

Example 11.13NCERT Example 11.13 · p. 191

A summer cooling bill

An air cooler rated 180 W runs for 10 hours a day through June (30 days). Electricity costs ₹7.50 per unit. What does it cost to run the cooler for the month?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Practice 11.7 (NCERT in-text questions, p. 192): questions with answers

  1. Short answerEasyNCERT in-text Questions p. 192, Q1

    Two appliances are plugged into the same 220 V supply. What decides which of them uses energy faster?

    Show solution

    AnswerThe current each draws (set by its resistance): P = VI, so the one drawing more current uses energy faster.

    The rate of energy use is the power, P=VI. With the same V, the appliance that draws the larger current, which is the one with the smaller resistance (P=V2/R), uses energy faster.

  2. NumericalEasyNCERT in-text Questions p. 192, Q2

    A water pump draws 4 A from a 220 V line and runs for 90 minutes.

    1. Find the pump's power.

      W
      Show answer

      Answer 880 W

      P=VI=220×4=880 W.
    2. How much energy does it use, in kW h?

      kW h
      Show answer

      Answer 1.32 kW h

      E=0.88 kW×1.5 h=1.32 kW h, which is 4.75×106 J.
Quick check
1 of 3
How many joules are there in 2.5 units of electrical energy?
J
Show answer

Answer 9000000 J

2.5×3.6×106=9.0×106 J.
Recap

Electricity: key points and summary

  1. Electric current is the rate of flow of charge, I=Q/t, measured in amperes; in metals electrons carry it, and conventional current is taken from + to − outside the source (Section 11.1).
  2. A cell or battery keeps up a potential difference, V=W/Q, measured in volts: 1 V=1 J/C (Section 11.2).
  3. An ammeter is joined in series; a voltmeter is joined in parallel across the two points (Section 11.2).
  4. Circuit diagrams use standard symbols; the long plate of a cell is its positive terminal (Section 11.3).
  5. Ohm's law: for a metal conductor at constant temperature V∝I, so V=IR; resistance is measured in ohms (Section 11.4).
  6. A rheostat changes the current by changing the resistance in the circuit (Section 11.4).
  7. R=ρl/A: resistance grows with length, falls with area and depends on the material through its resistivity, in Ω m (Section 11.5).
  8. In series the current is the same and the p.d.s add: Rs=R1+R2+R3 (Section 11.6.1).
  9. In parallel the p.d. is the same and the currents add: 1/Rp=1/R1+1/R2+1/R3, less than the smallest resistor (Section 11.6.2).
  10. Homes are wired in parallel so each appliance gets the full voltage and works on its own (Section 11.6.2).
  11. Joule's law of heating: H=VIt=I2Rt (Section 11.7).
  12. Heaters, filament bulbs and fuses use the heating effect; a fuse in series melts when the current is too large (Section 11.7.1).
  13. Electric power P=VI=I2R=V2/R, in watts; one unit is 1 kW h=3.6×106 J (Section 11.8).
Practice · 23 questions

Practice questions with answers (NCERT “Exercises”)

  1. Quick answerMediumNCERT Exercises Q1

    A uniform wire of resistance R is cut into four equal pieces, and the four pieces are joined in parallel. If the combination has resistance R′, the ratio R/R′ is:

    Show answer

    Answer (D) 16

    Each piece has R/4. Four equal resistors in parallel give a quarter of one: R′=R/16, so R/R′=16.
  2. Quick answerEasyNCERT Exercises Q2

    Which expression does not give the electrical energy used by a resistor in a time t?

    Show answer

    Answer (A) IR2t

    VIt, I2Rt and V2t/R are equal by Ohm's law. IR2t has the wrong units: it is not an energy.
  3. Quick answerMediumNCERT Exercises Q3

    An iron rated 800 W, 220 V is used when the supply drops to 176 V. If its resistance does not change, the power it uses is:

    Show answer

    Answer (D) 512 W

    P∝V2, and (176/220)2=0.82=0.64, so P=0.64×800=512 W. 640 W comes from forgetting to square.
  4. Quick answerHardNCERT Exercises Q4

    Three identical resistors are joined first in series and then in parallel, each time across the same p.d. for the same time. The ratio of the heat produced, series : parallel, is:

    Show answer

    Answer (A) 1:9

    Rs=3R and Rp=R/3. With V and t fixed, H∝1/R, so Hs:Hp=(R/3):3R=1:9.
  5. Short answerEasyNCERT Exercises Q5

    Asha wants the p.d. across a bulb, but she connects the voltmeter in series with it. What goes wrong, and how should the voltmeter be connected?

    Show solution

    AnswerIn series the voltmeter blocks the current, so the bulb goes out and the reading is wrong; it must be joined in parallel, across the two ends of the bulb.

    A voltmeter is built to be joined across two points. Put in series, it becomes part of the only path and its very large resistance almost stops the current: the bulb goes out and the meter reads nearly the whole battery p.d. Asha should join one voltmeter lead to each end of the bulb, in parallel with it.

  6. NumericalMediumNCERT Exercises Q6

    An aluminium wire has a diameter of 0.40 mm. Take its resistivity as 2.6×10−8 Ω m.

    1. What length of this wire has a resistance of 5.0 Ω?

      m
      Show answer

      Answer 24.2 m

      A=π(2.0×10−4)2=1.26×10−7 m2, so l=RA/ρ=5.0×1.26×10−7/(2.6×10−8)≈24 m.
    2. What would the resistance of the same length be if the diameter were halved?

      Ω
      Show answer

      Answer 20 Ω

      Halving the diameter makes the area a quarter, so the resistance becomes four times: 4×5.0=20 Ω.
  7. NumericalMediumNCERT Exercises Q7

    These readings were taken for a resistor. Plot V (upwards) against I and find the resistance from the slope of the best straight line through the origin.

    Table 11.9
    I (A)V (V)
    0.21.1
    0.42.3
    0.63.4
    0.84.7
    1.05.8
    Ω
    Show solution

    Plot the five points: they lie almost on one straight line through the origin, as Ohm's law predicts; the small scatter is experimental error.

    Slope of the best line =ΔV/ΔI≈5.8/1.0=5.8 Ω, which is the resistance.

  8. Quick answerEasyNCERT Exercises Q8

    A 9 V battery drives a current of 1.5 mA through an unknown resistor. Find its resistance in ohms.

    Ω
    Show answer

    Answer 6000 Ω

    R=V/I=9/(1.5×10−3)=6.0×103 Ω, that is 6 kΩ.
  9. NumericalEasyNCERT Exercises Q9

    A 6 V battery is joined in series with resistors of 0.5 Ω, 1.0 Ω, 1.5 Ω, 2.0 Ω and 10 Ω.

    1. What current flows through the 10 Ω resistor?

      A
      Show answer

      Answer 0.4 A

      Rs=0.5+1.0+1.5+2.0+10=15 Ω, so I=6/15=0.4 A through every resistor.
    2. What is the p.d. across the 10 Ω resistor?

      V
      Show answer

      Answer 4 V

      V=IR=0.4×10=4 V.
  10. Quick answerMediumNCERT Exercises Q10

    How many 330 Ω resistors must be joined in parallel to draw 4 A from a 220 V line?

    resistors
    Show answer

    Answer 6 resistors

    Needed: Rp=220/4=55 Ω. n equal resistors in parallel give 330/n, so n=330/55=6.
  11. DiagramMediumNCERT Exercises Q11

    You have three resistors of 4 Ω each. Show how to join all three to get a total of:

    1. 6 Ω

      Show solution

      AnswerOne 4 Ω in series with the other two in parallel.

      4∥4=2 Ω; then 4+2=6 Ω.

    2. 8/3 Ω (about 2.7 Ω)

      Show solution

      AnswerTwo in series (8 Ω), with the third in parallel across them.

      4+4=8 Ω; then 8∥4=32/12=8/3 Ω.

  12. Quick answerMediumNCERT Exercises Q12

    LED bulbs rated 12 W, 220 V are joined in parallel across a 220 V line that can safely carry at most 3 A. How many bulbs can be connected?

    bulbs
    Show answer

    Answer 55 bulbs

    One bulb draws I=P/V=12/220 A. The number is 3÷(12/220)=660/12=55.
  13. NumericalMediumNCERT Exercises Q13

    A room heater has two identical elements, P and Q, of 55 Ω each. A selector switch can put one element alone, both in series, or both in parallel across the 220 V mains.

    1. Current with one coil alone?

      A
      Show answer

      Answer 4 A

      I=220/55=4 A.
    2. Current with both coils in series?

      A
      Show answer

      Answer 2 A

      R=110 Ω, so I=220/110=2 A.
    3. Current with both coils in parallel?

      A
      Show answer

      Answer 8 A

      R=27.5 Ω, so I=220/27.5=8 A.
    4. Which setting heats fastest, and what is its power?

      Show solution

      AnswerParallel: 1760 W.

      Power P=VI: one coil 880 W, series 440 W, parallel 220×8=1760 W. The parallel setting heats fastest.

  14. NumericalMediumNCERT Exercises Q14

    In which of these two circuits does the 3 Ω resistor use more power, and how many times more?

    1. A 9 V battery with 3 Ω and 6 Ω resistors in series.

      W
      Show answer

      Answer 3 W

      I=9/9=1 A, so P=I2R=1×3=3 W.
    2. A 6 V battery with 3 Ω and 6 Ω resistors in parallel.

      W
      Show answer

      Answer 12 W

      The 3 Ω resistor has the full 6 V: P=V2/R=36/3=12 W, four times as much as in (a).
  15. Quick answerEasyNCERT Exercises Q15

    A 40 W tube light and a 70 W ceiling fan, both rated for 220 V, run in parallel on the 220 V mains. What total current do they draw?

    A
    Show answer

    Answer 0.5 A

    Total power 110 W, so I=P/V=110/220=0.5 A.
  16. Quick answerEasyNCERT Exercises Q16

    Which uses more energy: a 1500 W microwave oven run for 6 minutes, or a 75 W ceiling fan run for 3 hours?

    Show answer

    Answer (B) The ceiling fan

    Microwave: 1500×0.1 h=150 W h. Fan: 75×3=225 W h. The low-power fan uses more because it runs much longer.
  17. NumericalMediumNCERT Exercises Q17

    A geyser element of resistance 22 Ω draws 10 A from the mains for 30 minutes.

    1. At what rate is heat produced?

      W
      Show answer

      Answer 2200 W

      P=I2R=102×22=2200 W (joules per second).
    2. How much heat is produced in total?

      J
      Show answer

      Answer 3960000 J

      H=Pt=2200×1800=3.96×106 J, which is 1.1 kW h.
  18. Short answerMediumNCERT Exercises Q18 (a)–(d)

    Explain briefly:

    1. why bulb filaments are made of tungsten

      Show solution

      AnswerIt melts only at about 3400 °C, so it can glow white-hot without melting.

      A filament must reach well over 2500 ∘C to give light. Tungsten has a very high melting point (about 3400 ∘C) and stays strong when hot.

    2. why the elements of toasters and irons are made of an alloy, not a pure metal

      Show solution

      AnswerAlloys have higher resistivity and do not oxidise easily at high temperature.

      An alloy such as nichrome has a much higher resistivity than pure metals, so a short element gives a large resistance and plenty of heat; it also does not oxidise (burn) readily when red-hot.

    3. why appliances in a house are not joined in series

      Show solution

      AnswerIn series they would share the voltage, carry the same current and all stop if one fails.

      In series the mains voltage is shared, so no appliance gets its full 220 V; all must carry the same current though they need different currents; and one fault or switch-off breaks the circuit for all.

    4. how the resistance of a wire changes if its area of cross-section is doubled

      Show solution

      AnswerIt halves, because R ∝ 1/A.

      R=ρl/A, so with the same length and material, doubling A halves R.

  19. Short answerEasyNCERT Exercises Q18 (e)

    Why are copper and aluminium used for the wires that carry electricity across the country and into homes?

    Show solution

    AnswerTheir resistivities are very low, so little energy is lost as heat; aluminium is also light and cheaper.

    Copper and aluminium have some of the lowest resistivities, so long wires have small resistance and waste little energy as heat (I2R). Aluminium is also light and cheap, which suits overhead lines; copper is used in house wiring.

  20. Quick answerEasy

    Match each quantity with its SI unit.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • charge : coulomb
    • potential difference : volt
    • resistance : ohm
    • resistivity : ohm metre
    • power : watt
    C, V, Ω, Ω m and W. Resistivity has the extra metre because ρ=RA/l.
  21. Quick answerMedium

    Assertion (A): Two resistors joined in parallel always have an equivalent resistance smaller than either of them. Reason (R): In parallel, the reciprocals of the resistances add.

    Show answer

    Answer (A) Both A and R are true, and R explains A

    1/Rp=1/R1+1/R2 is larger than either 1/R1 or 1/R2, so Rp is smaller than either: R is true and explains A.
  22. Quick answerHard

    Assertion (A): A 100 W bulb glows more brightly than a 40 W bulb when both are joined in series across 220 V. Reason (R): In series, the bulb with the higher resistance uses the larger power.

    Show answer

    Answer (D) A is false but R is true

    In series both carry the same current, so P=I2R is larger for the larger resistance. The 40 W bulb has R=1210 Ω, more than the 100 W bulb's 484 Ω, so the 40 W bulb glows brighter. A is false, R is true.
  23. NumericalMedium

    A family's electricity meter reads 4512 units on 1 May and 4728 units on 1 June.

    1. How many units were used in May?

      kW h
      Show answer

      Answer 216 kW h

      4728−4512=216 units.
    2. What is the bill at ₹6.50 per unit?

      ₹
      Show answer

      Answer 1404 ₹

      216×6.50=1404, so ₹1404.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT in-text Questions p. 172, Q1Practice 11.1 (NCERT in-text questions, p. 172) · Q1
NCERT in-text Questions p. 172, Q2Practice 11.1 (NCERT in-text questions, p. 172) · Q2
NCERT in-text Questions p. 172, Q3Practice 11.1 (NCERT in-text questions, p. 172) · Q3
NCERT in-text Questions p. 174, Q1Practice 11.2 (NCERT in-text questions, p. 174) · Q1
NCERT in-text Questions p. 174, Q2Practice 11.2 (NCERT in-text questions, p. 174) · Q2
NCERT in-text Questions p. 174, Q3Practice 11.2 (NCERT in-text questions, p. 174) · Q3
NCERT in-text Questions p. 181, Q1Practice 11.3 (NCERT in-text questions, p. 181) · Q1
NCERT in-text Questions p. 181, Q2Practice 11.3 (NCERT in-text questions, p. 181) · Q2
NCERT in-text Questions p. 181, Q3Practice 11.3 (NCERT in-text questions, p. 181) · Q3
NCERT in-text Questions p. 181, Q4Practice 11.3 (NCERT in-text questions, p. 181) · Q4
NCERT in-text Questions p. 181, Q5Practice 11.3 (NCERT in-text questions, p. 181) · Q5
NCERT in-text Questions p. 185, Q1Practice 11.4 (NCERT in-text questions, p. 185) · Q1
NCERT in-text Questions p. 185, Q2Practice 11.4 (NCERT in-text questions, p. 185) · Q2
NCERT in-text Questions p. 188, Q1Practice 11.5 (NCERT in-text questions, p. 188) · Q1
NCERT in-text Questions p. 188, Q2Practice 11.5 (NCERT in-text questions, p. 188) · Q2
NCERT in-text Questions p. 188, Q3Practice 11.5 (NCERT in-text questions, p. 188) · Q3
NCERT in-text Questions p. 188, Q4Practice 11.5 (NCERT in-text questions, p. 188) · Q4
NCERT in-text Questions p. 188, Q5Practice 11.5 (NCERT in-text questions, p. 188) · Q5
NCERT in-text Questions p. 190, Q1Practice 11.6 (NCERT in-text questions, p. 190) · Q1
NCERT in-text Questions p. 190, Q2Practice 11.6 (NCERT in-text questions, p. 190) · Q2
NCERT in-text Questions p. 190, Q3Practice 11.6 (NCERT in-text questions, p. 190) · Q3
NCERT in-text Questions p. 192, Q1Practice 11.7 (NCERT in-text questions, p. 192) · Q1
NCERT in-text Questions p. 192, Q2Practice 11.7 (NCERT in-text questions, p. 192) · Q2
NCERT Exercises Q1Exercises · Q1
NCERT Exercises Q2Exercises · Q2
NCERT Exercises Q3Exercises · Q3
NCERT Exercises Q4Exercises · Q4
NCERT Exercises Q5Exercises · Q5
NCERT Exercises Q6Exercises · Q6
NCERT Exercises Q7Exercises · Q7
NCERT Exercises Q8Exercises · Q8
NCERT Exercises Q9Exercises · Q9
NCERT Exercises Q10Exercises · Q10
NCERT Exercises Q11Exercises · Q11
NCERT Exercises Q12Exercises · Q12
NCERT Exercises Q13Exercises · Q13
NCERT Exercises Q14Exercises · Q14
NCERT Exercises Q15Exercises · Q15
NCERT Exercises Q16Exercises · Q16
NCERT Exercises Q17Exercises · Q17
NCERT Exercises Q18 (a)–(d)Exercises · Q18
NCERT Exercises Q18 (e)Exercises · Q19

Quick check: MCQs on Electricity

  1. 11.1 Electric Current and Circuit3 questions
  2. 11.2 Electric Potential and Potential Difference3 questions
  3. 11.3 Circuit Diagram2 questions
  4. 11.4 Ohm's Law3 questions
  5. 11.5 Factors on which the Resistance of a Conductor Depends3 questions
  6. 11.6.1 Resistors in Series2 questions
  7. 11.6.2 Resistors in Parallel3 questions
  8. 11.7 Heating Effect of Electric Current2 questions
  9. 11.7.1 Practical Applications of Heating Effect of Electric Current3 questions
  10. 11.8 Electric Power3 questions
Glossary · 15

Key terms (English–Hindi)

ammeterऐमीटरaimeetar
A meter that measures current. It is joined in series, so the current being measured flows through it.First seen in the chapter →
electric circuitविद्युत परिपथvidyut paripath
An unbroken, closed conducting path that lets charge flow from one terminal of a source back to the other.First seen in the chapter →
electric currentविद्युत धाराvidyut dhara
The rate of flow of electric charge through a conductor, I=Q/t, measured in amperes (A).First seen in the chapter →
electric powerविद्युत शक्तिvidyut shakti
The rate at which electrical energy is used or supplied, P=VI, measured in watts (W).First seen in the chapter →
fuseफ़्यूज़fyooz
A short wire of low melting point joined in series with a circuit; it melts and breaks the circuit if the current gets too large.First seen in the chapter →
Joule's law of heatingजूल का तापन नियमjool ka tapan niyam
The heat produced in a resistor is H=I2Rt.First seen in the chapter →
kilowatt hourकिलोवाट घंटाkilowatt ghanta
The commercial unit of electrical energy (one unit): 1 kW used for 1 h, equal to 3.6×106 J.First seen in the chapter →
Ohm's lawओम का नियमom ka niyam
For a metal conductor at a steady temperature, the current is proportional to the potential difference: V=IR.First seen in the chapter →
parallel connectionपार्श्वक्रमparshvakram
Parts joined between the same two points, so each has the same potential difference across it.First seen in the chapter →
potential differenceविभवांतरvibhavantar
The work done to move one coulomb of charge between two points, V=W/Q, measured in volts (V).First seen in the chapter →
resistanceप्रतिरोधpratirodh
How strongly a conductor opposes the flow of charge, R=V/I, measured in ohms (Ω).First seen in the chapter →
resistivityप्रतिरोधकताpratirodhakta
A property of a material, ρ=RA/l, that fixes the resistance of a wire of given size; unit ohm metre (Ω m).First seen in the chapter →
rheostatधारा नियंत्रकdhara niyantrak
A variable resistor with a sliding contact, used to change the current without changing the source.First seen in the chapter →
series connectionश्रेणीक्रमshrenikram
Parts joined end to end in one chain, so the same current flows through each.First seen in the chapter →
voltmeterवोल्टमीटरvoltmeetar
A meter that measures potential difference. Its two leads are joined in parallel across the two points.First seen in the chapter →